Let $d\in\mathbb N$, $M\subseteq\mathbb R^d$, $k\in\{1,\ldots,d\}$, $\Omega$ be an open subset of $M$ and $\phi$ be a $C^1$-diffeomorphism from $\Omega$ onto an open subset of $\mathbb H^k:=\mathbb R^{k-1}\times[0,\infty)$.
We can show that, if $x\in\Omega$ and $u:=\phi(x)$, then $$T_x\:M=T_x\:\Omega={\rm D}\phi^{-1}(u)\mathbb R^k\tag1$$ and the pushforward $T_x(\phi)$ is an isomorphism between $T_x\:\Omega$ and $T_x\:\mathbb H^k=\mathbb R^k$.
By $(1)$, $T_x\:\Omega$ is $k$-dimensional. On the other hand, if $x\in\operatorname{Int}\Omega$ (the topological interior of $\Omega$), then $$B_\varepsilon(x)\subseteq\Omega\tag2$$ for some $\varepsilon>0$ and hence, if $v\in\mathbb R^d\setminus\{0\}$, $$\gamma(t):=x+tv\;\;\;\text{for }t\in\left(-\frac\varepsilon{\left\|v\right\|},\frac\varepsilon{\left\|v\right\|}\right)\tag3$$ is a $C^1$-curve on $\Omega$ through $x$ with $\gamma(0)=x$ and $\gamma'(0)=v$. This yields that $$T_x\:\Omega=\mathbb R^d\tag4.$$
Question 1: By $(2)$, $(4)$ is only possible if $k=d$. So, is there anything I'm missing or can we actually conclude that the topological interior $\operatorname{Int}\Omega$ (and hence $\operatorname{Int}M$) is empty, unless $k=d$?
Question 2: Assuming $k=d$, we know from $(2)$ that $T_x\:\Omega$ is $d$-dimensional for all $x\in\Omega$ (not necessarily $x\in\operatorname{Int}\Omega$). Can we conclude that $T_x\:\Omega=\mathbb R^d$?